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PHY53 Practice Exam Sol

# PHY53 Practice Exam Sol - PhysicsSS Fall 06 Exam 1 A ’...

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Unformatted text preview: PhysicsSS Fall 06 Exam 1 A ’ Name: Kg ‘2] Section: 1 certiﬁ) that 1 have abided by the rules and the spirit off/7e Duke Community Standard. PLEASE READ CAREFULLY BEFORE YOU START Do not forget to write your name and section number above. You must do all multiple choice Questions and two of the longer problems. If you attempt to do more than two of the longer problems, please, indicate below which two problems you want be counted. Remember that credit will only be given for solutions that Show your work. Do not just write down your result, but Show how you got it. No work shown = no credit. If numbers are required, use g = 10 m/sz. Record your MC answers (by letter A — E) here. No credit will be given, if you do not record your answer here! MCI C MCZ ID MC3 5 MC4 5 Check here, which of the two long problems should be counted: P1 P2 P3 P4 [To be ﬁlled in by graderz] Grading scores: MC P1 P2 P3 P4 Total: Multi )le Choice Problems: 5 Points Each l. A ball is thrown vertically upward at a certain initial speed and returns after 4.00 s to the same height from which it was thrown. Neglecting air resistance, how high above its launch height did the ball fly before falling back down? A) 10.0111. :1 , 5 V; 9:!- t=2,5. B 15.0 m. TX,“ V V0 ‘ cfwglem 5&3 o —-% 20.0111. 0 =Va... [05233.25 W go e. la Mfg D) 30.0 m. 5 E) 40.0 m. 2. How much work needs to be done by the motor to accelerate a car from rest to 30.0 m/s? The car’s mass is 1200 kg. [Neglect air drag. 1 id = 103 J] A) 54.0 k]. tiger AK: Wm) B) 72.0 H. Xi C) k]. " a- L a, va®5401<r AK“ Jaimie " 121W}? 3 12.4200 43°) = 5.‘toxiO§J E) 720 k]. 3. A rescue crew attempts to haul a ﬂood victim up to the helicopter by a rope that can withstand a tension force of at most 480 N. If the mass of the victim is 60.0 kg, what is the maximal acceleration with which she can be hauled up? W A) 4.00111/32. ' if; :— ‘tdoisl B) 6.00 111/32. C) 8.00 m/sz. D 12.0111/s2. I A . i = a W, :2 v- 69 v-ﬁ@ None, the rope always breaks under the victim’s weight. ‘2 %’ ON 4. The position of an object in an inertial reference frame is given by: x = a + b t+ 0 t2, where t denotes time. If a = "3 m, b = 2 m/s, and c = —l 111/52, you can conclude that: A) The object always moves in the positive x—direction. B) The object is accelerated in the positive x~direction. C) The object comes to rest for a brief moment at I = 2 s. D) The object initially moves in the negative x—direction, but then turns around at t = l s. The object accelerates in the negative x—direction and comes briefly to rest att=ls. V{5)=k>+2ct VUs) s 2?,2’4L3L.“ go (we): Zr. v02) 9 5: L3? Q, = 2c>-ng& The following problem is worth 7+7+6 = 20 points: P1. A member of the opposing basketball team passes the ball from a height of 2 m at an upward angle of 20° with respect to the horizontal and with a speed of 25 m/s. The ball misses the player for whom the pass was intended and falls to the ground. The Cameron crowd jeers. a) How long does it take from the moment the ball leaves the athlete’s hand for the ball to hit the ground? 13) How far does the ball fly in the horizontal direction before it reaches the ground? 0) At which angle with respect to the horizontal does the ball hit the ground? The followin0 problem is worth 6+8+6 = 20 points: P2. A science ﬁction writer wants to invent a ﬁctional planet which rotates so fast that the surface gravity at the equator just balances the centripetal force. Assume that the mass of the ﬁctional planet is 3.50 x 1024 kg and its radius is 4250 km, a) Draw the free—body diagrams of a person at the equator and of a person at 45° latitude (half way between the equator and the north pole) b) How long must the period of rotation of the planet be so that the inhabitants of this planet could get a “zero gravity” experience just by traveling to the equator? c) Find the magnitude of the static friction force needed to keep a person standing on the planet’s surface at 45° latitude from sliding toward the equator. Assume that the person’s mass is 60 kg. .9 M :5 ’ G fa) =3|§O Klb‘vf P VZ: F azilgkibém 2. Gi‘i a that w 3 {i3 ‘ if; M 23- 3 “(15(063 l=°AeZTiE—~=er——————L’/‘=‘( " s ‘9 Gm» a.e‘rxio“'a§.§oxro‘a‘t g 3... who“ 5 = sec-r035 (c) m; = mon r? cost”? (~ emit? e «mm L596?) 1? New“! A E~F" E” ~ Cw TF2 N " N3 f-r ‘ fie. N " K'emwwwt’ af l; =ma Ff? ~= w Wtwa‘f‘P we ‘59 Sal/#5" re “1 vi '9 “.3 v-t ,L G if a ‘é‘éo QSD5\$VRW 3.5mm) Lm (T? :2. 338A! The followino groblem is worth 4+8+8 = 20 points: P3. A truck pulls a trailer, which is connected to the truck by a cable, at constant speed up a road that rises at a 10° angle with respect to the horizontal. The truck’s mass is 1800 'ent of kinetic friction of the trailer kg, the trailer has a mass of 600 kg, and the coefﬁm is 0.12. a) Draw free-body diagrams for the truck and the trailer. Show and label all forces. (Do not forget to show the coordinate system you are using.) b) Find the tension force in the cable connecting the trailer and the truck. 0) How large must the coefficient of static friction between the wheels of the truck and the road be, so that the truck’s wheels will not slip? F}er a; ‘i‘WClL is static. (wheels cm mg! ) tiridx‘w 6? eruiler 4‘s, kinetic, (oi-it" msiSi‘m/tcﬁ; Nkrnml ﬂi‘c‘h‘om ,m ) Ft ‘ lit \$18 + Fm? 0 anti. 2 ‘ﬁaﬁl = “Mai/“5' Cd) 6 E}. = m3st rﬂk’wﬁ £059 "a mg, (st/t6 +jbtgwsﬁ) = 600%40f‘1Q5’QMlOO + @419 we 10°) = ’lt'ig-Kng/V (c) Fricﬁ'w ﬁrm 0; MM; CX”CMPW&A+) “5",. "" "FEEGS =0 -.- F; r mange = Ingroﬁi + [gook5.(o§:.sm 10° 17.5% = LL88 M03N Etméﬂ‘sa: =Mmﬁwse "'4 ﬂiéﬁﬁ m 3 lgoo (0%in (0° ...
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PHY53 Practice Exam Sol - PhysicsSS Fall 06 Exam 1 A ’...

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